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ANOTHER OPTIMIZATION PROBLEM

Heres a tricky one:

Problem 1. You need to design a cylindrical container consisting of two circularpieces (a top and bottom) and the lateral side piece. All of your material costs thesame amount: $ 1 per square meter. The lateral piece of the cylinder is formedby taking a rectangular piece and rolling it up. You therefore need to weld thecorresponding edges shut, and after the tube is made by welding one edge, the topand bottom must also be welded on. The cost for welding is $1 per meter. Thevolume of the cylinder is to be 2 cubic meters.

Due to manufacturing constraints, the radius of the cylinder must be at least onemeter. Find the cheapest cylinder you can build subject to all of these requirements.The volume of a cylinder of radius r and height h is r2h, and the total area is

2r2 + 2rh.

First, the volume is given and allows us to relate the height h of the cylinder tothe radius r:

2 = r2h, h =2

r2.

Next, the cost is given by $1 times the area, plus $1 times the welded length,which consists of the two circles at the top/bottom as well as one weld up thecylinder:

C(r, h) = (2r2 + 2rh) + (2(2r) + h).

Plugging in our known relation between h and r allows us to express this as afunction of just one variable:

C(r) = 2r2 +4

r+ 4r +

2

r2.

Taking a derivative, we have

C(r) = 4r 4

r2+ 4

4

r3.

Solving for zero would be quite a chore, but recall that we were requires to haver 1. Note that

C(1) = 4 4 > 0,

So as we increase the radius to be larger than one, the cost is increasing. Note alsothat

C(r) = 4 + 8r3

+ 12r4

,

which is always positive. So C(r) is always increasing, which means that (sinceC(1) > 0) that for all r > 1 we must have C(r) > 0: the cost is always increasing.The cheapest container is therefore found at the endpoint value of r = 1. The costof this cylinder is

C(1) = 10 + 2.

That one was tricky, but a nice little argument that I thought you should see atleast once. Here is a problem which is more boilerplate.

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2 ANOTHER OPTIMIZATION PROBLEM

Problem 2. You wish to build a cylinder to hold magic seawater. The heightof your cylinder must be equal to the diameter of the top and bottom. . . for safety

reasons. All materials cost the same, and there is no welding cost. If your con-tainer holds a total volume of V, then you can sell the container for $ V1/3. Whatdimensions for your container will maximize your profit?

Well, weve been told that h = 2r explicitly, so the volume V is given by

V(r) = r2h = 2r3,

so our gross income G(r) will be

G(r) = V(r)1/3

= r (2)1/3.

On the other hand, our costs C(r) are given by the area of the container, which is

C(r) = 2r2 + 2rh = 2r2 + 4r2 = 6r2.

So our profit P(r) is given by our income minus our costs:

P(r) = G(r) C(r) = r (2)1/3 6r2.

We must have r 0, but otherwise we have no real restrictions. So we take aderivative, remembering the chain rule:

P(r) = (2)1/3 12r.

Setting the derivative equal to zero, we need to solve for r:

0 =(2)1/3 12r

12r =(2)1/3

r =(2)1/3

12

To check that this is indeed a maximum for our profit, we will use the secondderivative test:C(r) = 12,

which is negative for all values of r. Therefore this value is indeed a maximum.